4
$\begingroup$

Let $F_2$ be the free group of rank 2, and let $G$ be a finite 2-generated group. To any such $G$, we may associate the set of surjections $$G(F_2) := Surj(F_2,G)$$ which admits a natural action of $Aut(F_2)$. Furthermore, any surjection $p : G\twoheadrightarrow G'$ induces a natural map $$p_* : \underbrace{Surj(F_2,G)}_{G(F_2)}\longrightarrow \underbrace{Surj(F_2,G')}_{G'(F_2)}$$ which by Gaschutz' lemma is a surjection of $Aut(F_2)$-sets.

I wish to understand if:

  1. Every $Aut(F_2)$-equivariant automorphism $G(F_2)\rightarrow G(F_2)$ actually comes from an automorphism $\alpha\in Aut(G)$.
  2. Every $Aut(F_2)$-equivariant surjection $G(F_2)\rightarrow G'(F_2)$ actually comes from a surjection $G\rightarrow G'$.
  3. Same as (1), but now looking at $Aut(F_2)$-orbits. Ie, can you have an $Aut(F_2)$-equivariant automorphism of an $Aut(F_2)$-orbit of $G(F_2)$ that doesn't come from an automorphism $\alpha\in Aut(G)$?
  4. Same as (2), but looking at $Aut(F_2)$-orbits. Ie, can you have an $Aut(F_2)$-equivariant surjection from an $Aut(F_2)$-orbit of $G(F_2)$ to an orbit of $G'(F_2)$ that doesn't come from a surjection $G\rightarrow G'$?
  5. Are (1),(2),(3), or (4) true if we restrict the groups $G$ we consider? For example, if we restrict $G$ to only nonabelian finite simple groups (possibly letting $G'$ be anything)?

I mentioned the Yoneda embedding in the title because philosophically, I am asking if $G(F_2)$, as an $Aut(F_2)$-set, is "equivalent" to the functor $Surj(*,G)$.

Question (1) is probably the simplest, and I believe it is equivalent to asking:

(1a) Does the action of $Aut(F_2)$ on $\Omega_G := G(F_2)/Aut(G)$ have trivial centralizer inside the symmetric group on $\Omega_G$?

If $G$ is a nonabelian finite simple group, let $N_G := |\Omega_G|$. Then $N_G$ is called the $G$-rank of $F_2$, and in fact by a Goursat lemma argument, the product of the $N_G$ distinct representatives of $\Omega_G$ yields a surjection $F_2\twoheadrightarrow G^{N_G}$ with characteristic kernel. From this, if we fix an embedding $F_2\rightarrow\widehat{F_2}$ into the profinite completion, Gaschutz' lemma tells us that $Aut(\widehat{F_2})$ acts as the full symmetric group $S_{\Omega_G}$ on $\Omega_G$, which of course has trivial centralizer in $S_{\Omega_G}$. Of course, $Aut(F_2)$ is much much smaller than $Aut(\widehat{F_2})$, but on the other hand (1a) only requires that the centralizer be trivial.

Has this question been considered? I would also appreciate any relevant references.

$\endgroup$
  • $\begingroup$ 1,2 are false (as you probably anticipated when formulating 3). Actually they fail for every non-abelian simple group. The reason is that, when $G$ is simple, the action of $Aut(G)$ on every $Aut(F_2)$-orbit in $G(F_2)$ is faithful. Therefore it is enough to show that there are at least 2 such orbits. Namely, given a conjugacy class $c$ in $G$, the subset of $G(F_2)$ consisting of those generating pairs $(x,y)$ such that $[x,y]\in c^{\pm 1}$, is $Aut(F_2)$-invariant. Since $[x,y]^{\pm 1}$ can achieve several conjugacy classes (would need an argument...), this shows this non-transitivity. $\endgroup$ – YCor Nov 1 '16 at 3:21
  • 1
    $\begingroup$ Yes but it's enough to consider one nontrivial inner automorphism $u$ of $G$, then fix some conjugacy class, and define an $Aut(F_2)$-equivariant permutation $\Phi_{u,c}$ of $G(F_2)$ as being the identity outside $G(F_2)_c$, and coinciding with composition by $u$ on $G(F_2)_c$. Here $G(F_2)_c$ consists of those generating pairs $(x,y)$ such that $[x,y]\in c^{\pm 1}$. From remarks of my previous comment, $\Phi_{u,c}$ does not coincide with composition by an automorphism of $G$. $\endgroup$ – YCor Nov 1 '16 at 3:37
  • 1
    $\begingroup$ Sounds intriguing. I'd be curious to understand, say for the simple group of order 60, what is the image of $Aut(F_2)$ in the group of permutations of a given orbit of a generating pair. Anyway a useful keyword would be Nielsen equivalence. $\endgroup$ – YCor Nov 1 '16 at 5:10
  • 1
    $\begingroup$ @DerekHolt yes, it's part of what I'm asking. A refined question is to describe how $Aut(F_2)$ acts on these sets of 9 and 10 elements (roughly, what's the image of $Aut(F_2)$ in $S_9$ and $S_{10}$?) $\endgroup$ – YCor Nov 1 '16 at 7:17
  • 1
    $\begingroup$ Ah, interesting. This implies that the centralizer of this action is trivial... so at the level of the action on a given $Aut(F_2)$-orbit of generating pairs (i.e., before modding out by the action of $Aut(G)$, this suggests (I haven't checked) that the action of the centralizer of the $Aut(F_2)$-action preserves the $Aut(G)$-orbits (hum, there's a little issue that $Aut(G)$ could permute $Aut(F_2)$-orbits, so I'm a bit vague). Anyway this makes me optimistic to establish (3) (at least in this case of $Alt_5$). $\endgroup$ – YCor Nov 1 '16 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.